Astone is dropped from the top of 50 m high tower simultaneously another stone is thrown upward with a speed of 20 m/s . calculate the time at which both the stones cross each other

'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.

For the stone dropped from the 50m tower:

H = +50 - (1/2) G T²

For the stone tossed upward from the ground:

H = +20T - (1/2) G T²

When the stones' paths cross, their Heights are equal.

50 - (1/2) G T² = 20T - (1/2) G T²

Wow !  Look at that !  Add (1/2) G T² to each side of that equation,
and all we have left is:

50 = 20T  Isn't that incredible ? ! ?

Divide each side by 20 :

2.5 = T

The stones meet in the air 2.5 seconds after the drop/toss.

I want to see something: 
What is their height, and what is the tossed stone doing, when they meet ?

Their height is  +50 - (1/2) G T² = 19.375 meters

The speed of the tossed stone is  +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it.  The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward